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Operations Scheduling Krishna Murari

128opertaions Scheduling

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Page 1: 128opertaions Scheduling

Operations Scheduling

Krishna Murari

Page 2: 128opertaions Scheduling

Scheduling

Scheduling is process of setting up operations processing time so that job is completed by the time they are due.

The main objective of scheduling to provide the best service to the customer through efficient use of resources.

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Purpose of Scheduling

To help the firm to maximise customer satisfaction

To Minimize service delays To Enable the firm to allocate the

capacity to meet customers’ demands on time.

To reduce cost and maximise profit To create flexibility to accommodate

variation in customer demand

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Manufacturing Execution Systems

Continuous – Assembly type – large number of units of homogenous product is produced

Intermittent – production of variety of product one at a time or in batches (custom made as per customer’s requirement)

Combination – neither strictly continuous nor intermittent

Sequence in which waiting jobs are processed is critical to efficiency and effectiveness of the intermittent system

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Work Center

A work center is an area in a business in which productive resources are organized and work is completed

Can be a single machine, a group of machines, or an area where a particular type of work is done

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Scheduling Methods

Infinite loading : jobs are assigned to work centers without regard to the work center’s capacity as if it’s capacity is infinite. Gantt load charts and visual load profiles and assignment algorithm are used to evaluate loading and assigning the jobs.

Finite loading : A scheduling procedure that assigns jobs into work centers and dermines their starting and completion dates by considering the work center’s capacity. Work center’s capacity is allocated unit by unit by simulating job starting and completion time.

Selection of scheduling method depends on volume of production, capacity of work center and nature of operations

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Scheduling Methods

Forward scheduling : Determining the start and finish times for the waiting jobs by assigning them to the earliest available time slots at the work centre. (Used in job shops). Forward scheduling is simple in use and it gets job done in shortest lead time but has excessive inventory as jobs get accumulated at various work centers as they get completed before the availability of next work center.

Backward scheduling : Determining the start and finish times for the waiting jobs by assigning them to the latest available time slots at the work centre to enable to complete each job just when it is due. (Used in assy). It minimizes the inventory since job is not completed till it must go to next work center on its routing.

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Scheduling Methods

2 jobs are to be processed on 2 machines. The route sheets is given below. Both jobs should be ready in 8 hours. Develop schedule using forward and backward scheduling.

Job X route Sheet Job Y route sheetRouting sequence

Machine Processing Time Hrs

Routing sequence

Machine Processing Time Hrs

1 1 2 1 1 2

2 2 3 2 2 3

3 1 1

Total 6 5

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Scheduling MethodsForward Scheduling

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X 1

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X 2

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Scheduling MethodsBackward Scheduling

Job X route Sheet Job Y route sheetRouting sequence

Machine Processing Time Hrs

Routing sequence

Machine Processing Time Hrs

1 1 2 1 1 2

2 2 3 2 2 3

3 1 1

Total 6 5

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Scheduling Methods

Backward Scheduling

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Y 1

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Y 2

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Typical Scheduling and Control Functions

Allocating orders, equipment, and personnel

Determining the sequence of order performance

Initiating performance of the scheduled work

Shop-floor control

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Work-Center Scheduling Objectives

Meet due dates

Minimize lead time

Minimize setup time or cost

Minimize work-in-process inventory

Maximize machine utilization

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Scheduling Activities

Routing : it is specifications of work flow. It explains the sequence of operations and processes to be followed in order to produce a particular product. It defines the what to do and where and how to do. Route sheets give these details.

Loading : It is assigning specific job to each work centre for planned period. For loading capacity limitation of each machine is to be considered.

Dispatching : It is the final act of releasing the job orders to the worker to go ahead with the production process. In this, operation manager releases the job orders as per the planned sequence and then controls the process for effective implementation of schedule.

Dispatching rules called as priority rules are used in scheduling the production activities.

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Priority Rules for Job Sequencing 1. First-come, first-served (FCFS) or first-in, first served :

Jobs are processed in the order of their arrival, all jobs are treated as equally important for example – petrol filling

2. Shortest operating time (SOT) or shortest processing time (SPT) : priorities is given based on the shortest processing time of the jobs. This is used when firm wants that maximum no. of jobs should be completed.

3. Earliest due date first (DDate) : The jobs are prioritized according to their earliest due dates. Dispatching is done in such a way that the one with earliest due date is dispatched first, the next earliest next and so on.

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Priority Rules for Job Sequencing 4. Slack time remaining (STR) first : In this method, the

operating manager calculate the slack time of each job i.e. the difference between the time remaining in the due date and the processing time required. Job with the shortest slack time is dispatched first.

5. Slack time remaining per operation (STR/OP) : Orders with shortest STR/OP are run first as follows:

Time remaining before due date

Remaining Processing time _

STR/OP =

Number of Remaining operations

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Priority Rules for Job Sequencing6. Critical ratio (CR) : This is calculated as the difference

between the due date and the current date by the work remaining. Order with the smallest CR are run first.

7. Last come, first served (LCFS) : This happens frequently by default. As orders arrive, they are placed on the top of the stack and operator using picks up from top and run.

8. Random order or whim : the supervisor or operator selects whichever he feels like running.

9. Longest processing time

10. Start date-due date minus normal lead time.

remaining days of Number

date) Current-date (DueCR

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Example of Job Sequencing: First-Come First-Served

Jobs (in order Processing Due Date Flow Timeof arrival) Time (days) (days hence) (days)

A 4 5 4B 7 10 11C 3 6 14D 1 4 15

Answer: FCFS Schedule

Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)

A 4 5B 7 10C 3 6D 1 4

Suppose you have the four jobs to the right arrive for processing on one machine

Suppose you have the four jobs to the right arrive for processing on one machine

What is the FCFS schedule?What is the FCFS schedule?

No, Jobs B, C, and D are going to be late

No, Jobs B, C, and D are going to be late

Do all the jobs get done on time?Do all the jobs get done on time?

Delay A-0, B-1, C-8 and D – 11 days

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Example of Job Sequencing: Shortest Operating Time (SOT or SPT)

Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)

A 4 5B 7 10C 3 6D 1 4

Answer: Shortest Operating Time Schedule

Jobs (in order Processing Due Date Flow Timeof arrival) Time (days) (days hence) (days)

D 1 4 1C 3 6 4A 4 5 8B 7 10 15

Suppose you have the four jobs to the right arrive for processing on one machine

Suppose you have the four jobs to the right arrive for processing on one machine

What is the SOT schedule?What is the SOT schedule?

No, Jobs A and B are going to be late

No, Jobs A and B are going to be late

Do all the jobs get done on time?Do all the jobs get done on time?

Delay D-0, C-0, A-3 and B -5 days

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Example of Job Sequencing: Earliest Due Date First (DDate)

Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)

A 4 5B 7 10C 3 6D 1 4

Answer: Earliest Due Date First

Jobs (in order Processing Due Date Flow Timeof arrival) Time (days) (days hence) (days)

D 1 4 1A 4 5 5C 3 6 8B 7 10 15

Suppose you have the four jobs to the right arrive for processing on one machine

Suppose you have the four jobs to the right arrive for processing on one machine

What is the earliest due date first schedule?

What is the earliest due date first schedule?

No, Jobs C and B are going to be late

No, Jobs C and B are going to be late

Do all the jobs get done on time?Do all the jobs get done on time?

Delay D-0, A-0, C-2 and B -5 days

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Job Sequencing: Critical Ratio Method

Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)

A 4 5B 7 10C 3 6D 1 4

Suppose you have the four jobs to the right arrive for processing on one machine

Suppose you have the four jobs to the right arrive for processing on one machine

What is the CR schedule?What is the CR schedule?

No, but since there is a three-way tie, only the first job or two will be on time

No, but since there is a three-way tie, only the first job or two will be on time

In order to do this schedule the CR’s have be calculated for each job. If we let today be Day 1 and allow a total of 15 days to do the work. The resulting CR’s and order schedule are:CR(A)=(5-4)/15=0.06 (Do this job first)CR(B)=(10-7)/15=0.20 (Do this job second, tied with C and D)CR(C)=(6-3)/15=0.20 (Do this job second, tied with B and D)CR(D)=(4-1)/15=0.20 (Do this job second, tied with B and C)

Do all the jobs get done on time?Do all the jobs get done on time?

remaining days of Number

date) Current-date (DueCR

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Job Sequencing:Last-Come First-Served (LCFS)

Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)

A 4 5B 7 10C 3 6D 1 4

Answer: Last-Come First-Served ScheduleJobs (in order Processing Due Date Flow Time

of arrival) Time (days) (days hence) (days)D 1 4 1C 3 6 4B 7 10 11A 4 5 15

No, Jobs B and A are going to be late

No, Jobs B and A are going to be late

Suppose you have the four jobs to the right arrive for processing on one machine

Suppose you have the four jobs to the right arrive for processing on one machine

What is the LCFS schedule?What is the LCFS schedule?Do all the jobs get done on time?Do all the jobs get done on time?

Delay D-0, C-0, B-1 and A -10 days

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Slack Time Remaining

No, Jobs D,C and B are going to be late

No, Jobs D,C and B are going to be late

Suppose you have the four jobs to the right arrive for processing on one machine

Suppose you have the four jobs to the right arrive for processing on one machine

What is the STR schedule?What is the STR schedule?

Do all the jobs get done on time?Do all the jobs get done on time?

Delay A-0, D-1, C-2 and B-5 days

Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)

A 4 5B 7 10C 3 6D 1 4

Slack Time for A, B,C and D are (5-4), (10-7), (6-3) and (4-1) , B , C and D have same slack time but D has less processing time so D will be before C &B and sequence will be A,D,C,B

Job sequence

Processing time

Due date Time Flow

A 4 5 4

D 1 4 5

C 3 6 8

B 7 10 15

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Scheduling Techniques

Many techniques are used by operations manager. Important techniques are :

Gantt chart

Job sequencing rule

Queuing theory

Critical ratio

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Gantt Charts

It was developed by Henry L. Gantt. It is a graphical display of the duration of a set of activities.

Gantt chart: Used as a tool to monitor the progress of work and to view the load on workstations. The chart takes two basic forms: (1) the job or activity

progress chart, and (2) the workstation chart.

The Gantt progress chart graphically displays the current status of each job or activity relative to its scheduled completion date.

The Gantt workstation chart shows the load on the workstations and the nonproductive time.

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Gantt Progress ChartGantt Progress Chart

Plymouth

Ford

Pontiac

Job 4/11 6/11 7/11 8/11 9/11 12/115/111/11 2/11 3/11

Current Current date date

Scheduled activity timeActual progress

Start activity

Finish activity

Nonproductive time

Gantt Progress Chart for an Auto Parts Company

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Gantt Workstation Chart Gantt Workstation Chart

Gantt Workstation Chart for Hospital Operating Rooms

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Johnson’s Job Sequencing Rules

Sequencing of jobs is important for scheduling of jobs. It helps in minimizing the processing time and maximizing the operations efficiency. It also reduces the operations cost over a period of time.

When jobs are processed in multistage, it is necessary to sequence in such a way that idle time is reduced.

Johnson's rules developed by Johnson and Bellman, are utilized to minimized the total time span required for completing the given jobs.

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Johnson’s Rule for Two stage Production

If a firm has to perform “n” jobs on two machines A and B in the order AB. Expected time for these jobs are A1, A2 , A3….. An on machine A and B1, B2, B3….. Bn in machine B. Following steps are followed:

1. Identify least processing time in A1, A2,….An and B1, B2….Bn. If there is tie , select either of the processing time.

2. If the smallest processing time is Ar (rth job in machine A then place it at the beginning of the sequence. And if it is Bs (sth job on machine B), then place it at the end of the sequence.

3. If there is tie for the least processing time on machine A, any of the jobs can be placed at the beginning of the sequence. In case of tie at machine B, any of the jobs can be placed at the end of the sequence.

Contd…

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Johnson’s Rule for Two stage Production 4. Identify the next least processing time and repeat the above

step. The process is continued till all the jobs are assigned in a sequence. The sequence obtained is called as optimum sequence.

5. Once the time sequence is found, the total elapse time and idle time on machine A and B can be calculated by the following formula.

Total Elapse time = time between start of the first job – end time of last job.

Idle time on machine A = time of completion of last job on machine B in optimum sequence- time of completion of last job on machine A in optimum sequence

Idle time on machine B = Time taken by Machine A in completing first job in sequence - ∑ [ ( time when k th job starts on machine B) – (time when k-1 th Job finishes on B)

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Example of Job Sequencing: Johnson’s Rule (Part 1)

Suppose you have the following five jobs with time requirements in two stages of production. What is the job sequence using Johnson’s Rule?

Suppose you have the following five jobs with time requirements in two stages of production. What is the job sequence using Johnson’s Rule?

Time in HoursJobs Stage 1 Stage 2 A 1.50 1.25 B 2.00 3.00 C 2.50 2.00 D 1.00 2.00

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Example of Job Sequencing: Johnson’s Rule (Part 2)

First, select the job with the smallest time in either stage.

That is Job D with the smallest time in the first stage. Place that job as early as possible in the unfilled job sequence below.

Drop D out, select the next smallest time (Job A), and place it 4th in the job sequence as it is at stage 2

Drop A out, select the next smallest time. There is a tie in two stages for two different jobs. In this case, place the job with the smallest time in the first stage as early as possible in the unfilled job sequence. Then place the job with the smallest time in the second stage as late as possible in the unfilled sequence.

Job Sequence 1 2 3 4

Job Assigned D A B C

Time in HoursJobs Stage 1 Stage 2 A 1.50 1.25 B 2.00 3.00 C 2.50 2.00 D 1.00 2.00

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Problem : A firm produces six type of shoes pairs and manufacturing of shoes pairs require processing on two machines A and B. The processing time for each shoe pair on both machines (in Hrs) are given below:

Shoe pair type

1 2 3 4 5 6

Machine A

35 105 55 25 95 105

Machine B

75 100 95 65 35 20

Determine an optimum sequence so that the total elapse time is minimum. Also calculate idle time on machine A and Machine B

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Solution: stage:1 Least processing time is 20 for job 6 at machine B hence it will come at the end of sequence i.e. at 6th place

Shoe pair type

6

Machine A

105

Machine B

20

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Solution: stage 2 Next least process time is 25 hrs for shoe pair type 4 on machine A hence it will be placed at the beginning of sequence

Shoe pair type

4 6

Machine A

25 105

Machine B

60 20

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Solution: stage: 3 Next least process time is 30 hrs for shoe pair type 1 on machine A and 30 hrs for type 5 pair on machine B so type 1 is scheduled nest to type 4 on machine A and type 5 shoe pair before type 6 pairs shown below

Shoe pair type

4 1 5 6

Machine A

25 35 95 105

Machine B

65 75 35 20

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Solution: stage: 4 Next least process time is 55 hrs for shoe pair type 3 on machine , hence it will be sequenced after shoe pair 1

Shoe pair type

4 1 3 5 6

Machine A

25 35 55 95 105

Machine B

65 75 95 35 20

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Solution: stage:5 last left our pair 2 will be sequenced after pair 3

Shoe pair type

4 1 3 2 5 6

Machine A

25 35 55 105 95 105

Machine B

65 75 95 100 35 20

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Solution: stage:6Now we develop the schedule of each job in optimum sequence as shown below

Job sequence

Machine A Machine B

Time in Processing time

Time out

Time in Processing time

Time out

4 0 25 25 25 65 90

1 25 35 60 90 75 165

3 60 55 115 165 95 260

2 115 105 220 260 100 360

5 220 95 315 360 35 395

6 315 105 420 420 20 440

Note : Time- in for machine will be time-out of the job from machine A or time-out of previous job from machine B which ever is bigger

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Solution: stage:7 We can calculate elapse time and idle time as follows

Total elapse time : 440 hours

Idle time for machine A = 440-420 = 20 hrs

Idle time for machine B = Time taken by machine A to complete first job 25 hrs + [ (90-90)+(165-165) + (260-260) +(360-360) + (420-395)] = 50 hrs.

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Johnson’s Rule for Three Stage Production

If the expected processing time for n-jobs on 3 machines A, B and C are A1,A2,A3…..An; B1, B2, B3….Bn and C1,C2,C3…..Cn

Stage -1 : check that Johnson rule can be used:

Any one of the following conditions should be satisfied

i) The smallest processing time on machine A should be greater than or equal to the largest processing time on machine B

ii) The smallest processing time on machine C should be greater than or equal to the largest processing time of machine B

Contd..

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Johnson’s Rule for Three Stage Production

Stage -2 : Assume two fictitious machines G and H

Stage -3 : Find out corresponding time for these machines by they following formula

Gi = Ai +Bi

Hi = Bi + Ci

Stage -3 : solve the problem like 2 machines and n jobs with order of GH.

The resulting optimum sequence will also become the optimum sequence for the problem of 3 machines and n jobs.

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Example : Johnson’s Rule for Three Stage Production

A firm is involved in five types of jobs, each must process in 3 machines A, B, C in the order ABC. The processing time (in hours) for each job on the three machines is given below

Job Processing time

Ai Bi Ci

1 18 10 8

2 19 12 18

3 12 5 16

4 16 6 14

5 21 9 10

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Example : Three Stage ProductionFirst we check that condition to apply Johnson’s rule is satisfied. Minimum processing time at machine A is 12 hrs. and maximum processing time at B is 12hrs , it satisfied the condition , hence the rule can be used. Now we use Gi=Ai+Bi andHi=Bi+Ci, the table will be as given below

Job Processing time

Gi = Ai+Bi Hi=Bi+Ci

1 28 18

2 31 30

3 17 21

4 22 20

5 30 19

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Example : Three Stage ProductionThe optimum sequence will be as follows

Minimum processing time of job 3 is 17 hrs in G hence it will be first in sequence i.e. job 3 will be first in sequence. Now, in remaining

Jobs, job 1 in machine H is lowest processing time 18 hrs, hence it will be the last in sequence.

Now , in left out jobs, job 5 has lowest processing time in machine H, hence, it will come last but one job i.e. before job 1.

Job 4 has the lowest time in machine H so it will be before the job 5 and job 2 will be in left out slot i.e. after job 3.

Hence sequence will be 3,2,4,5,1. The schedule will be as given in table in next slide.

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Example : Three Stage Production

Job sequence

Machine A Machine B Machine C

Time in

Processing time

Time Out

Time in

Processing time

Time Out

Time in

Processing time

Time Out

3 0 12 12 12 5 17 17 16 33

2 12 19 31 31 12 43 43 18 61

4 31 16 47 47 6 53 61 14 75

5 47 21 68 68 9 77 77 10 87

1 68 18 86 86 10 96 96 8 104

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Example : Three Stage Production

Total operation time with optimum sequence is 104 hrs

Idle time on machine A = 104 -86 =14 hrs.

Idle time on machine B= 12 + [ (31-17)+ (47-43)+ (68-53) + (77-86) + (104 -96) ]= 62 hrs.

Idle time on machine C = 17 +[ (43-33)+ (61-61)+ (77-75) +(96-87) = 38 hrs.

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Johnson’s Rule for n jobs and m-machines

If n-jobs 1,2,3…,n on M machines A1,A2,A3,….Am are to ebsceduled. The processing time of job 1 on m machines are A11, A12, A13…..A1m ; processing time for job 2 are A21, A22, A23…..A2m and processing time for job n are An1, An2, An3….Anm.

Stage -1 : check that Johnson rule can be used:

Any one of the following conditions should be satisfied

i) The smallest processing time on machine A1 should be greater than or equal to the largest processing time on machines A2, A3, A4…. Am-1

ii) The smallest processing time on machine Am should be greater than or equal to the largest processing time of machines A2, A3, A4…. Am-1

Contd..

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Johnson’s Rule for n jobs on m machines

Stage -2 : Assume two fictitious machines G and H

Stage -3 : Find out corresponding time for these machines by they following formula

Gi = Ai 1+Ai2 + Ai3…..+Aim-1

Hi = Ai2 + Ai3…..+Aim-1 +Aim

Stage -3 : solve the problem like 2 machines and n jobs with order of GH.

The resulting optimum sequence will also become the optimum sequence for the problem of m machines and n jobs.

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A firm is involved in 4 types of jobs, each must process in 4 machines P,Q,R and S in the order PQRS. The processing time (in hours) for each job on the 4 machines is given below

Job Processing time

P Q R S

A 10 7 5 9

B 9 6 4 7

C 8 4 2 6

D 12 8 3 9

Johnson’s Rule for n jobs on m machines

Determine the optimum sequence , elapse time and idle time on each machine

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First we check that condition to apply Johnson’s rule is satisfied. Minimum processing time at machine P is 8 hrs. and maximum processing time at Q and R are 8 and 5 , it satisfied the condition , hence the rule can be used. Now we use Gi=A1+A2+A3 A2+A3+A4 the table will be as given below

Job Processing time

Gi = A1+A2+A3 Hi=A2+A3+A4

A 22 21

B 19 17

C 14 12

D 23 20

Johnson’s Rule for n jobs on m machines

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The optimum sequence will be as follows

Minimum processing time of job C is 12 hrs in H hence it will be last in sequence.

Now in remaining Jobs, job B in machine H is lowest processing time 17 hrs, hence it will be the last but one in sequence.

Now , in left out jobs, job D has lowest processing time in machine H hence, it will come before job B

Hence sequence will be A,D,B and C. The schedule will be as given in table in next slide.

Johnson’s Rule for n jobs on m machines

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Job Machine

P Q R S

A 0-10 10-17 17-22 22-31

D 10-22 22-30 30-33 33-42

B 22-31 31-37 37-41 42-49

C 31-39 39-43 43-45 49-55

Johnson’s Rule for n jobs on m machines

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Total operation time with optimum sequence is 55 hrs

Idle time on machine P = 55-39 =16 hrs.

Idle time on machine Q = 10+(22-17)+(31-30)+(39-37)+(55-43) = 30 hrs.

Idle time on machine R = 17+(30-22)+(37-33)+(43-41)+(55-45)= 41 hrs

Idle time on machine S =22+(33-31)+(42-42)+(49-49)=24hrs

Johnson’s Rule for n jobs on m machines

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Personnel Scheduling in Services

Issues :

Customer’s direct interaction

High variability in demand

Two operations:

Front of the house

Back of the house

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Personnel Scheduling in Services

Scheduling consecutive days off – 5 days week working

Scheduling daily work times

Scheduling hourly work times

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Personnel Scheduling in Services

Methods to meet customer demand under limited availability of service facilities

Appointment system – customer arrival time can be controlled

Reservation system – like hotel rooms

Strategic product pricing – to adjust the shift in demand like higher price at peak hours – electricity at high price summer